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''method of images''
 
''method of images''
  
A method in potential theory for solving certain boundary value problems for partial differential equations in a domain <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623001.png" />, where the fulfillment of the boundary conditions on the boundary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623002.png" /> is achieved by the choice of additional field sources situated outside <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623003.png" /> and called image sources.
+
A method in potential theory for solving certain boundary value problems for partial differential equations in a domain $  D $,  
 +
where the fulfillment of the boundary conditions on the boundary $  \partial  D = \Gamma $
 +
is achieved by the choice of additional field sources situated outside $  D $
 +
and called image sources.
 +
 
 +
The greatest development of the mapping method occurred in electrostatics. Suppose, for example, that it is required to solve the [[Dirichlet problem|Dirichlet problem]] for the [[Poisson equation|Poisson equation]]  $  \Delta u = - 2 \pi \rho ( x, y) $
 +
in the half-plane  $  D = \{ {( x, y) } : {y > 0,  - \infty < x < + \infty } \} $
 +
with a given function  $  \psi ( x) $
 +
on the boundary  $  \Gamma = \{ {( x, y) } : {y = 0,  - \infty < x < + \infty } \} $,
 +
that is, it is required to find the potential of electrostatic charges of density  $  \rho ( x, y) $
 +
situated in  $  D $,
 +
with the condition that the potential  $  \psi ( x) $
 +
is fixed on  $  \Gamma $.  
 +
It is known that in order to solve this problem it suffices to know the [[Green function|Green function]]  $  G ( x, y;  x _ {0} , y _ {0} ) $
 +
representing the potential of a unit point charge at the point  $  ( x _ {0} , y _ {0} ) \in D $
 +
when the boundary  $  \Gamma $
 +
is earthed, that is,  $  G ( x, 0;  x _ {0} , y _ {0} ) = 0 $.  
 +
The solution  $  u ( x, y) $
 +
of the original problem is expressed as follows in terms of  $  G ( x, y;  x _ {0} , y _ {0} ) $:
  
The greatest development of the mapping method occurred in electrostatics. Suppose, for example, that it is required to solve the [[Dirichlet problem|Dirichlet problem]] for the [[Poisson equation|Poisson equation]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623004.png" /> in the half-plane <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623005.png" /> with a given function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623006.png" /> on the boundary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623007.png" />, that is, it is required to find the potential of electrostatic charges of density <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623008.png" /> situated in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m0623009.png" />, with the condition that the potential <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230010.png" /> is fixed on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230011.png" />. It is known that in order to solve this problem it suffices to know the [[Green function|Green function]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230012.png" /> representing the potential of a unit point charge at the point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230013.png" /> when the boundary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230014.png" /> is earthed, that is, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230015.png" />. The solution <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230016.png" /> of the original problem is expressed as follows in terms of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230017.png" />:
+
$$ \tag{1 }
 +
u ( x, y)  = \
 +
{\int\limits \int\limits } _ { D }
 +
\rho ( x _ {0} , y _ {0} )
 +
G ( x, y;  x _ {0} , y _ {0} ) \
 +
dx _ {0}  dy _ {0} +
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230018.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
+
$$
 +
+
 +
{
 +
\frac{1}{2 \pi }
 +
} \int\limits _  \Gamma  \psi ( x _ {0} )
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230019.png" /></td> </tr></table>
+
\frac{\partial  G ( x, y; x _ {0} , 0) }{\partial  y _ {0} }
 +
  dx _ {0} .
 +
$$
  
When there is no boundary, the potential of a point charge can be written as a fundamental solution <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230020.png" /> of the [[Laplace equation|Laplace equation]]. Adding a negative unit charge image at the point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230021.png" /> and forming the sum of the potentials of these two charges, one obtains the desired Green function:
+
When there is no boundary, the potential of a point charge can be written as a fundamental solution $  \mathop{\rm ln} ( 1/ \sqrt {( x - x _ {0} )  ^ {2} + ( y - y _ {0} )  ^ {2} } ) $
 +
of the [[Laplace equation|Laplace equation]]. Adding a negative unit charge image at the point $  ( x _ {0} , - y _ {0} ) $
 +
and forming the sum of the potentials of these two charges, one obtains the desired Green function:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230022.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
+
$$ \tag{2 }
 +
G ( x, y; x _ {0} , y _ {0} ) =
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230023.png" /></td> </tr></table>
+
$$
 +
= \
 +
\mathop{\rm ln} {
 +
\frac{1}{\sqrt {( x - x _ {0} )  ^ {2} + ( y - y _ {0} )  ^ {2} } }
 +
} -  \mathop{\rm ln} {
 +
\frac{1}{\sqrt {( x - x _ {0} )  ^ {2} + ( y + y _ {0} )  ^ {2} } }
 +
} .
 +
$$
  
In the case of a strip <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230024.png" />, on reflecting a unit charge at <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230025.png" /> with respect to the lines <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230026.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230027.png" /> one gets an infinite sequence of charges <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230028.png" /> at the points <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230029.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230030.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230031.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230032.png" /> respectively. The Green function in this case can be expressed in the form of an infinite series of potentials of point charges.
+
In the case of a strip $  D = \{ {( x, y) } : {0 < y < b,  - \infty < x < + \infty } \} $,  
 +
on reflecting a unit charge at $  ( x _ {0} , y _ {0} ) \in D $
 +
with respect to the lines $  y = 0 $
 +
and $  y = b $
 +
one gets an infinite sequence of charges $  - 1, - 1, + 1, + 1 \dots $
 +
at the points $  ( x _ {0} , - y _ {0} ) $,
 +
$  ( x _ {0} , 2b - y _ {0} ) $,
 +
$  ( x _ {0} , - 2b + y _ {0} ) $,  
 +
$  ( x _ {0} , 2b + y _ {0} ) \dots $
 +
respectively. The Green function in this case can be expressed in the form of an infinite series of potentials of point charges.
  
For a domain <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230033.png" /> in the form of a disc, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230034.png" />, the image of a unit charge at <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230035.png" /> is a negative unit charge at <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230036.png" />, which is the image of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230037.png" /> under [[Inversion|inversion]] relative to the circle <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230038.png" />.
+
For a domain $  D $
 +
in the form of a disc, $  D = \{ {( \rho , \phi ) } : {0 \leq  \rho < a,  0 \leq  \phi < 2 \pi } \} $,  
 +
the image of a unit charge at $  ( \rho _ {0} , \phi _ {0} ) $
 +
is a negative unit charge at $  ( a  ^ {2} / \rho _ {0} , \phi _ {0} ) $,  
 +
which is the image of $  ( \rho _ {0} , \phi _ {0} ) $
 +
under [[Inversion|inversion]] relative to the circle $  \rho = a $.
  
 
Other configurations of boundaries are also possible, consisting of lines and circles, and the solution is obtained by constructing the corresponding sequence of charge images.
 
Other configurations of boundaries are also possible, consisting of lines and circles, and the solution is obtained by constructing the corresponding sequence of charge images.
  
For the solution of the Dirichlet problem for the Poisson equation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230039.png" /> in the half-space
+
For the solution of the Dirichlet problem for the Poisson equation $  \Delta u = - 4 \pi \rho ( x, y, z) $
 +
in the half-space
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230040.png" /></td> </tr></table>
+
$$
 +
= \{ {( x, y, z) } : {z > 0, - \infty < x, y < + \infty } \}
 +
,
 +
$$
  
on reflecting the unit charge at <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230041.png" /> with respect to the plane <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230042.png" />, instead of (2) one obtains the formula
+
on reflecting the unit charge at $  ( x _ {0} , y _ {0} , z _ {0} ) \in D $
 +
with respect to the plane $  z = 0 $,  
 +
instead of (2) one obtains the formula
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230043.png" /></td> </tr></table>
+
$$
 +
G ( x, y, z; x _ {0} , y _ {0} , z _ {0} ) =
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230044.png" /></td> </tr></table>
+
$$
 +
= \
 +
{
 +
\frac{1}{\sqrt {( x - x _ {0} )  ^ {2} + ( y - y _ {0} )  ^ {2} + ( z - z _ {0} )  ^ {2} } }
 +
} +
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230045.png" /></td> </tr></table>
+
$$
 +
- {
 +
\frac{1}{\sqrt {( x - x _ {0} )  ^ {2} + ( y - y _ {0} )  ^ {2} + ( z + z _ {0} )  ^ {2} } }
 +
} .
 +
$$
  
In the case of a ball <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230046.png" /> it is necessary to apply the [[Kelvin transformation|Kelvin transformation]], and the image of the unit charge at the point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230047.png" /> is a charge of magnitude <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230048.png" /> at the point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230049.png" />, which is the image of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230050.png" /> under inversion with respect to the sphere <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230051.png" />. From this one obtains that if a solution of the Poisson equation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230052.png" /> is known in some domain <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230053.png" />, then the function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230054.png" /> gives a solution of the Poisson equation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230055.png" /> with density <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230056.png" /> in the domain <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230057.png" /> that is the image of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230058.png" /> under inversion with respect to the sphere <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230059.png" />. In this form the mapping method is sometimes called the method of inversion. In applying the method of inversion it is necessary to pay attention to the fact that the boundary conditions are also transformed.
+
In the case of a ball $  D = \{ {( r, \theta , \phi ) } : {0 \leq  r < a,  0 \leq  \theta \leq  \pi,  0 \leq  \phi < 2 \pi } \} $
 +
it is necessary to apply the [[Kelvin transformation|Kelvin transformation]], and the image of the unit charge at the point $  ( r _ {0} , \theta _ {0} , \phi _ {0} ) \in D $
 +
is a charge of magnitude $  - a/r _ {0} $
 +
at the point $  ( a  ^ {2} /r _ {0} , \theta _ {0} , \phi _ {0} ) $,  
 +
which is the image of $  ( r _ {0} , \theta _ {0} , \phi _ {0} ) $
 +
under inversion with respect to the sphere $  r = a $.  
 +
From this one obtains that if a solution of the Poisson equation $  \Delta u = - 4 \pi \rho ( r, \theta , \phi ) $
 +
is known in some domain $  D $,  
 +
then the function $  v ( r, \theta , \phi ) = ( a/r) u ( a  ^ {2} /r, \theta , \phi ) $
 +
gives a solution of the Poisson equation $  \Delta u = - 4 \pi \rho  ^  \prime  ( r, \theta , \phi ) $
 +
with density $  \rho  ^  \prime  ( r, \theta , \phi ) = ( a/r)  ^ {5} \rho ( a  ^ {2} /r, \theta , \phi ) $
 +
in the domain $  D  ^  \prime  $
 +
that is the image of $  D $
 +
under inversion with respect to the sphere $  r = a $.  
 +
In this form the mapping method is sometimes called the method of inversion. In applying the method of inversion it is necessary to pay attention to the fact that the boundary conditions are also transformed.
  
 
For more complicated domains in space, whose boundaries consist of some planes or spheres, it is also possible to apply infinite sequences of charge images. In conjunction with passage to limits, where one or more sources go to infinity, the mapping method makes it possible to solve complicated problems, such as the determination of the potential of an electrostatic field in the case of a conducting ball placed in a field that is homogeneous at infinity.
 
For more complicated domains in space, whose boundaries consist of some planes or spheres, it is also possible to apply infinite sequences of charge images. In conjunction with passage to limits, where one or more sources go to infinity, the mapping method makes it possible to solve complicated problems, such as the determination of the potential of an electrostatic field in the case of a conducting ball placed in a field that is homogeneous at infinity.
  
In the case of the [[Helmholtz equation|Helmholtz equation]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230060.png" />, the mapping method is applicable only for domains bounded by lines or for domains in space bounded by planes and uses the corresponding fundamental solutions <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230061.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230062.png" /> or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230063.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m062/m062300/m06230064.png" />.
+
In the case of the [[Helmholtz equation|Helmholtz equation]] $  \Delta u + k  ^ {2} u = 0 $,  
 +
the mapping method is applicable only for domains bounded by lines or for domains in space bounded by planes and uses the corresponding fundamental solutions $  H _ {0}  ^ {(} 1) ( k \rho ) $,  
 +
$  H _ {0}  ^ {(} 2) ( k \rho ) $
 +
or $  e  ^ {ikr} /r $,  
 +
$  e  ^ {-} ikr /r $.
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  G.A. Grinberg,  "A collection of problems in the mathematical theory of electrical and magnetic phenomena" , Moscow-Leningrad  (1948)  (In Russian)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  J.D. Jackson,  "Classical electrodynamics" , Wiley  (1962)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  N.E. Kochin,  I.A. Kibel',  N.V. Roze,  "Theoretical hydrodynamics" , Interscience  (1964)  (Translated from Russian)</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  W.R. Smythe,  "Static and dynamic electricity" , McGraw-Hill  (1950)</TD></TR></table>
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  G.A. Grinberg,  "A collection of problems in the mathematical theory of electrical and magnetic phenomena" , Moscow-Leningrad  (1948)  (In Russian)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  J.D. Jackson,  "Classical electrodynamics" , Wiley  (1962)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  N.E. Kochin,  I.A. Kibel',  N.V. Roze,  "Theoretical hydrodynamics" , Interscience  (1964)  (Translated from Russian)</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  W.R. Smythe,  "Static and dynamic electricity" , McGraw-Hill  (1950)</TD></TR></table>

Latest revision as of 07:59, 6 June 2020


method of images

A method in potential theory for solving certain boundary value problems for partial differential equations in a domain $ D $, where the fulfillment of the boundary conditions on the boundary $ \partial D = \Gamma $ is achieved by the choice of additional field sources situated outside $ D $ and called image sources.

The greatest development of the mapping method occurred in electrostatics. Suppose, for example, that it is required to solve the Dirichlet problem for the Poisson equation $ \Delta u = - 2 \pi \rho ( x, y) $ in the half-plane $ D = \{ {( x, y) } : {y > 0, - \infty < x < + \infty } \} $ with a given function $ \psi ( x) $ on the boundary $ \Gamma = \{ {( x, y) } : {y = 0, - \infty < x < + \infty } \} $, that is, it is required to find the potential of electrostatic charges of density $ \rho ( x, y) $ situated in $ D $, with the condition that the potential $ \psi ( x) $ is fixed on $ \Gamma $. It is known that in order to solve this problem it suffices to know the Green function $ G ( x, y; x _ {0} , y _ {0} ) $ representing the potential of a unit point charge at the point $ ( x _ {0} , y _ {0} ) \in D $ when the boundary $ \Gamma $ is earthed, that is, $ G ( x, 0; x _ {0} , y _ {0} ) = 0 $. The solution $ u ( x, y) $ of the original problem is expressed as follows in terms of $ G ( x, y; x _ {0} , y _ {0} ) $:

$$ \tag{1 } u ( x, y) = \ {\int\limits \int\limits } _ { D } \rho ( x _ {0} , y _ {0} ) G ( x, y; x _ {0} , y _ {0} ) \ dx _ {0} dy _ {0} + $$

$$ + { \frac{1}{2 \pi } } \int\limits _ \Gamma \psi ( x _ {0} ) \frac{\partial G ( x, y; x _ {0} , 0) }{\partial y _ {0} } dx _ {0} . $$

When there is no boundary, the potential of a point charge can be written as a fundamental solution $ \mathop{\rm ln} ( 1/ \sqrt {( x - x _ {0} ) ^ {2} + ( y - y _ {0} ) ^ {2} } ) $ of the Laplace equation. Adding a negative unit charge image at the point $ ( x _ {0} , - y _ {0} ) $ and forming the sum of the potentials of these two charges, one obtains the desired Green function:

$$ \tag{2 } G ( x, y; x _ {0} , y _ {0} ) = $$

$$ = \ \mathop{\rm ln} { \frac{1}{\sqrt {( x - x _ {0} ) ^ {2} + ( y - y _ {0} ) ^ {2} } } } - \mathop{\rm ln} { \frac{1}{\sqrt {( x - x _ {0} ) ^ {2} + ( y + y _ {0} ) ^ {2} } } } . $$

In the case of a strip $ D = \{ {( x, y) } : {0 < y < b, - \infty < x < + \infty } \} $, on reflecting a unit charge at $ ( x _ {0} , y _ {0} ) \in D $ with respect to the lines $ y = 0 $ and $ y = b $ one gets an infinite sequence of charges $ - 1, - 1, + 1, + 1 \dots $ at the points $ ( x _ {0} , - y _ {0} ) $, $ ( x _ {0} , 2b - y _ {0} ) $, $ ( x _ {0} , - 2b + y _ {0} ) $, $ ( x _ {0} , 2b + y _ {0} ) \dots $ respectively. The Green function in this case can be expressed in the form of an infinite series of potentials of point charges.

For a domain $ D $ in the form of a disc, $ D = \{ {( \rho , \phi ) } : {0 \leq \rho < a, 0 \leq \phi < 2 \pi } \} $, the image of a unit charge at $ ( \rho _ {0} , \phi _ {0} ) $ is a negative unit charge at $ ( a ^ {2} / \rho _ {0} , \phi _ {0} ) $, which is the image of $ ( \rho _ {0} , \phi _ {0} ) $ under inversion relative to the circle $ \rho = a $.

Other configurations of boundaries are also possible, consisting of lines and circles, and the solution is obtained by constructing the corresponding sequence of charge images.

For the solution of the Dirichlet problem for the Poisson equation $ \Delta u = - 4 \pi \rho ( x, y, z) $ in the half-space

$$ D = \{ {( x, y, z) } : {z > 0, - \infty < x, y < + \infty } \} , $$

on reflecting the unit charge at $ ( x _ {0} , y _ {0} , z _ {0} ) \in D $ with respect to the plane $ z = 0 $, instead of (2) one obtains the formula

$$ G ( x, y, z; x _ {0} , y _ {0} , z _ {0} ) = $$

$$ = \ { \frac{1}{\sqrt {( x - x _ {0} ) ^ {2} + ( y - y _ {0} ) ^ {2} + ( z - z _ {0} ) ^ {2} } } } + $$

$$ - { \frac{1}{\sqrt {( x - x _ {0} ) ^ {2} + ( y - y _ {0} ) ^ {2} + ( z + z _ {0} ) ^ {2} } } } . $$

In the case of a ball $ D = \{ {( r, \theta , \phi ) } : {0 \leq r < a, 0 \leq \theta \leq \pi, 0 \leq \phi < 2 \pi } \} $ it is necessary to apply the Kelvin transformation, and the image of the unit charge at the point $ ( r _ {0} , \theta _ {0} , \phi _ {0} ) \in D $ is a charge of magnitude $ - a/r _ {0} $ at the point $ ( a ^ {2} /r _ {0} , \theta _ {0} , \phi _ {0} ) $, which is the image of $ ( r _ {0} , \theta _ {0} , \phi _ {0} ) $ under inversion with respect to the sphere $ r = a $. From this one obtains that if a solution of the Poisson equation $ \Delta u = - 4 \pi \rho ( r, \theta , \phi ) $ is known in some domain $ D $, then the function $ v ( r, \theta , \phi ) = ( a/r) u ( a ^ {2} /r, \theta , \phi ) $ gives a solution of the Poisson equation $ \Delta u = - 4 \pi \rho ^ \prime ( r, \theta , \phi ) $ with density $ \rho ^ \prime ( r, \theta , \phi ) = ( a/r) ^ {5} \rho ( a ^ {2} /r, \theta , \phi ) $ in the domain $ D ^ \prime $ that is the image of $ D $ under inversion with respect to the sphere $ r = a $. In this form the mapping method is sometimes called the method of inversion. In applying the method of inversion it is necessary to pay attention to the fact that the boundary conditions are also transformed.

For more complicated domains in space, whose boundaries consist of some planes or spheres, it is also possible to apply infinite sequences of charge images. In conjunction with passage to limits, where one or more sources go to infinity, the mapping method makes it possible to solve complicated problems, such as the determination of the potential of an electrostatic field in the case of a conducting ball placed in a field that is homogeneous at infinity.

In the case of the Helmholtz equation $ \Delta u + k ^ {2} u = 0 $, the mapping method is applicable only for domains bounded by lines or for domains in space bounded by planes and uses the corresponding fundamental solutions $ H _ {0} ^ {(} 1) ( k \rho ) $, $ H _ {0} ^ {(} 2) ( k \rho ) $ or $ e ^ {ikr} /r $, $ e ^ {-} ikr /r $.

References

[1] G.A. Grinberg, "A collection of problems in the mathematical theory of electrical and magnetic phenomena" , Moscow-Leningrad (1948) (In Russian)
[2] J.D. Jackson, "Classical electrodynamics" , Wiley (1962)
[3] N.E. Kochin, I.A. Kibel', N.V. Roze, "Theoretical hydrodynamics" , Interscience (1964) (Translated from Russian)
[4] W.R. Smythe, "Static and dynamic electricity" , McGraw-Hill (1950)
How to Cite This Entry:
Mapping method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Mapping_method&oldid=12837
This article was adapted from an original article by E.D. Solomentsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article